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sussBest ResponseYou've already chosen the best response.0
it was supposed to x to the power logylogz y to the power logzlogx z to the power logxlogy
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
\(\huge x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}\) like this ?
 one year ago

sussBest ResponseYou've already chosen the best response.0
ah thank god..yes like that
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
consider \(\huge \log[x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}]\) can u simplify that ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
you should get that =0 = ln 1
 one year ago

sussBest ResponseYou've already chosen the best response.0
i dont understand y u multiplied the whole by log????
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
else, nothing can't be really done to that expression
 one year ago

sussBest ResponseYou've already chosen the best response.0
i have done that problem in this way \[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]
 one year ago

sussBest ResponseYou've already chosen the best response.0
and i cudnt figure out the next step..... :(
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
i know, i tried that also....since i could not proceed, i thought of taking log....
 one year ago

sussBest ResponseYou've already chosen the best response.0
if u dont mind,can u show me the simplification process...??? :)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
\(\huge \log[x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}] \\\large =\log[x^{\log y\log z}]+\log[y^{\log z\log x}]+\log[z^{\log x\log y}]\) got this ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
now use \( \\ \log A^{B}=B\log A\)
 one year ago

sussBest ResponseYou've already chosen the best response.0
oh sorry man i jus lost connection leeme check it out
 one year ago

sussBest ResponseYou've already chosen the best response.0
y did u keep the + sign ???
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
because \(\log AB=\log A+\log B \\ \)
 one year ago

sussBest ResponseYou've already chosen the best response.0
ah finally got it .....thx a ton man
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
sure ? u getting that = 0 ? then write 0=log 1
 one year ago

sussBest ResponseYou've already chosen the best response.0
yes 100%..i hav to use formula of log ab bak ther ......i really got it man
 one year ago
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